You have found the following ages (in years) of all 6 snakes at your local zoo: $ 6,\enspace 13,\enspace 22,\enspace 7,\enspace 9,\enspace 8$ What is the average age of the snakes at your zoo? What is the variance? You may round your answers to the nearest tenth.
Answer: Because we have data for all 6 snakes at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{6 + 13 + 22 + 7 + 9 + 8}{{6}} = {10.8\text{ years old}} $ Find the squared deviations from the mean for each snake. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $6$ years $-4.8$ years $23.04$ years $^2$ $13$ years $2.2$ years $4.84$ years $^2$ $22$ years $11.2$ years $125.44$ years $^2$ $7$ years $-3.8$ years $14.44$ years $^2$ $9$ years $-1.8$ years $3.24$ years $^2$ $8$ years $-2.8$ years $7.84$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{23.04} + {4.84} + {125.44} + {14.44} + {3.24} + {7.84}} {{6}} $ $ {\sigma^2} = \dfrac{{178.84}}{{6}} = {29.81\text{ years}^2} $ The average snake at the zoo is 10.8 years old. The population variance is 29.81 years $^2$.